Answer to Question #126533 in Mechanics | Relativity for Mwansa Kunda

Question

Two objects are connected by a light string passing over a frictionless pulley. m1=5.00kg, m2=3.00kg, h=4.00m
The object of mass 5kg was released from rest. Using the principle of conservation of energy,
Determine;
i. The speed of the 3.0kg object just as the 5.0kg object hits the ground.
ii. The maximum height to which the 3.0kg object rises.

Answer 1

5 kg body locates 4 meters above ground. After it released it will go down because it heavier than 3 kg body. Energy conservation principle says:

$E_{p}=E_{k}$

(1) $m\times a\times h = \frac{m\times V^2}{2}$

where m - mass of body (5 kg), a - acceleration (not equals to gravitational acceleration in our case because we have 3kg body on other side), h - height (4 meters), V - velocity of body near ground.

acceleration could be found:

$F_{result} = F_{g5} - F_{g3}= m_{5}\times g - m_{3} \times g;$

$F_{result} = 5\times 9.8 - 3\times 9.8 = 49 - 29.4 = 19.6N$;

$F_{result}=m_{5}\times a;$

$a=\frac{F_{result}}{m_{}}=\frac{19.6}{5}=3.92 m/s^2$

so, returning to formula (1) we can find V of body 5kg and body 3 kg (they a the same in this moment)

$V=\sqrt{\frac{m_{5}\times a \times h \times 2}{m_{5}}}=\sqrt{a \times h \times 2}=\sqrt{3.92\times 4\times 2};$

$V=5.6m/s$

Ansver for first question speed is 5.6 m/s

After 5 kg body hit ground, body 3kg will locate on height 8 m and continue move upward for some time, because it has some speed. To find maximum height which body 3 will have, let's use the same principle:

$m_{3}\times g\times h=\frac{m_{3}\times V^2}{2};$

$h=\frac{V^2}{2\times g}$

this time we use g because only force that affects body 3 kg is gravity.

$h=\frac{5.6^2}{2\times 9.8}=1.6m$

So maximal height of body 3 kg is 8 + 1.6 = 9.6m